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A function $f:E\to \mathbb{R}$ has uniform continuity if for every $\epsilon >0$ there is a $\delta >0$ such that, for all $x,y\in E$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. The same $\delta$ must work everywhere in $E$.

The product need not be uniformly continuous. Both $f(x)=x$ and $g(x)=x$ are Lipschitz continuous, hence uniformly continuous on $\mathbb{R}$, but their product is $x^2$. For $x_n=n$ and $y_n=n+1/n$, the input separation tends to zero whileThis contradicts uniform continuity. The issue is captured by products of unbounded uniformly continuous functions; bounded factors would give a uniform product estimate.

A uniformly continuous function on $(0,1)$ is bounded. Take $\delta$ for output tolerance one and cover $(0,1)$ by finitely many intervals of radius less than $\delta$, centred at points $c_1,\dots ,c_m$ in $(0,1)$. Every $x$ is within $\delta$ of some $c_j$, so $|f(x)|\le 1+{\max }_j|f(c_j)|$. This proves the general principle that a uniformly continuous function on a totally bounded set is bounded.

The reciprocal-cosine function is not uniformly continuous. TakeBoth sequences lie in $(0,1)$, $|x_n-y_n|\to 0$, but the two function values are $1$ and $-1$. Their difference is always two, violating the sequential criterion for uniform continuity.